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For a real diode the current as function of the potential difference between its terminals (for big enough voltage) is given by:

$$I=A e^{\frac{eV-E}{\eta k_B T}} \tag{1}$$

So in the case of LEDs can I interpret $A$ as the current it starts emitting light?

My reasoning is that E is gap energy and eV is the total energy administrated, when they nullify $I=A$, after this point the rest of the energy is converted into light and heat, so A must be the current we see the LED starts shinning, but I need a second opinion.

Details: I'm supposed to use (1) to determine Planck's constant, the approximation (1) works well for V>1V as I was told (the voltage LEDs start emitting light is around 1.5V), I measured different values of voltage and current in different LEDs, as well as their maximum emitted wavelength, and the temperature where the experiment was done, to find $E=\dfrac{hc}{\lambda} (2)$ I thought of using least squares, rewriting (1) as:

$$eV=\eta k_B T\ln{(I/A)}+E$$

Then $\eta$ is the slope of the curve, E the interception with the eV-axis, I know the current and voltage, so I can use least squares to find $\eta$ of each LED and the corresponding E, then I can use (2) to determine h for each one and then take their mean value as the constant to compare with the real one, but A is what I don't know, is it the current the LED starts emitting electromagnetic radiation?.

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  • $\begingroup$ A is an area factor. Two identical diodes in parallel will have twice the current at the same voltage, so the "compound diode" will have twice the A. Two identical LEDs will also emit twice the amount of light at the same voltage compared to one. Ideally the light emitted by an LED should be proportional to I, but that's not the case, so you have to be careful if that plays a role in your application. $\endgroup$ – CuriousOne Jun 26 '15 at 1:47
  • $\begingroup$ I'm trying to measure Planck's constant with that formula and least squares so I need to know if A can be measured as the current when it starts emitting EMR, if A depends on the area, the material and temperature is also important but I'm more interested in how to measure it the simplest way $\endgroup$ – user162485 Jun 26 '15 at 2:43
  • $\begingroup$ I think the short answer is "no". $A$ has the units of current, but it is hard to attribute a meaning to its magnitude. What about the voltage applied? I would be curious to know more about your experiment. $\endgroup$ – Floris Jun 26 '15 at 3:21
  • $\begingroup$ You don't need to measure A but only measure the activation voltage as a function of wavelength. The activation voltage does not depend on A. A only tells you how much current will flow/how much light will be emitted when light will be emitted. At the voltage where no light will be emitted, at all, A does not matter (no light times 2 is still no light). $\endgroup$ – CuriousOne Jun 26 '15 at 3:22
  • $\begingroup$ I think what probably bothers you is that the equation never makes $I=0$, right? That's because you have been neglecting a term. The equation is the simplified diode equation. The complete diode equation is $I_d=I_0[e^{qV_d/nkT}-1]$, which does go to zero for a zero voltage. There is also a non-zero probability for light emission from an LED due to thermal excitation below the threshold, but you can really forget about that effect, it makes a mV size shift for your purposes. The exponential drops so fast that one can use a finite threshold and still get away with it. $\endgroup$ – CuriousOne Jun 26 '15 at 3:42

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